Reno giveaway

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I just received a promotion from one of the Reno casinos about a dice-based giveaway. You throw five dice and your prize is based on the poker hand generated. A large straight (1-5 or 2-6) pays $5,000. Is this the correct formula for the probability of hitting such a payout --

There are indeed 6^5 ways = 7776 ways to throw the dice. You are looking for combinations of 2, 3, 4, 5, 6.

Assume that the 1st die is any number (be it a 1 - 6). The 2nd die must be different from the 1st die. That has a 5/6 shot. The third die must be different from the 1st two that has a 4/6 shot. Then you have 3/6 and 2/6.

So, that leaves 5/6*4/6*3/6*2/6, giving you a .092593. But of these, only a fraction are straights. You have 12346 12356 12456 13456 and 12345 and 23456. So, multiply by 1/3 to get .030864.

However, if this is the Harrah's promotion, the prize structure is as folllows:

Prizes: Five of A Kind = $5,000Four of A Kind = $1,000Large Straight = $750Full House = $500Small Straight = $2503 of a kind = $150Two Pairs = $100One Pair = $50Less than one pair wins another roll until prize is earned

There are indeed 6^5 ways = 7776 ways to throw the dice. You are looking for combinations of 2, 3, 4, 5, 6.

Assume that the 1st die is any number (be it a 1 - 6). The 2nd die must be different from the 1st die. That has a 5/6 shot. The third die must be different from the 1st two that has a 4/6 shot. Then you have 3/6 and 2/6.

So, that leaves 5/6*4/6*3/6*2/6, giving you a .092593. But of these, only a fraction are straights. You have 12346 12356 12456 13456 and 12345 and 23456. So, multiply by 1/3 to get .030864.

However, if this is the Harrah's promotion, the prize structure is as folllows:

Prizes: Five of A Kind = $5,000Four of A Kind = $1,000Large Straight = $750Full House = $500Small Straight = $2503 of a kind = $150Two Pairs = $100One Pair = $50Less than one pair wins another roll until prize is earned

By golly, that's an advantage play if I ever saw one! Can they run out of dice?